[bkcooper]

This is false. A single particle can still be described by the maximum entropy probability distribution given its average energy . . .

I'm not sure that construction makes sense. For a single particle not interacting with anything, imposing the maxent constraint on average energy just sets the energy of the particle and there's no room left for a distribution. As soon as its interacting with another system you're good, but then the temperature that you arrive it is a characteristic of the bath, not the particle itself.

[asafira]

how does the average energy set the energy of the particle?

To me, for any configuration of the particle x, the average energy of the particle is given by

sum over all configurations of p(x) E(x)

and there's the probability distribution over configurations.

[alecst]

I think what the OPs mean is that in the classical case, the lone, noninteracting particle has a fixed, definite energy and not an average energy. There are no other configurations to consider, besides the set of points that the particle may be in space, but that is irrelevant to the energy and temperature. Maybe what you're saying is true if you consider a single quantum particle, I'm not sure.

[asafira]

I don't think invoking quantum mechanics is necessary here honestly.

Let's say I flip a coin; most people would model the outcome as a p = 1/2 chance of heads and p = 1/2 chance of tails. This is fairly standard, of course, but I think it illustrates my point.

Of course, in reality there is only 1 outcome that's going to happen, but literally the best model of the system we can come up with, given our ignorance, is that it's 50/50 heads and tails.

So, what I am saying is that given that we know relatively little about the particle --- what it's average energy is --- we can come up with a probability distribution over states with that average energy that represents our "best guess" or "least biased" probability assignments.

[evanb]

I think the issue is that, assuming the particle is alone or only elastically bouncing off of its container, E(x) is a constant E in which case (let's assume no internal degrees of freedom) the configuration contains only position and orientation information, rather than any information about a spread in energies.

edit: sniped by alecst.

[posterboy]

As would be the case for particles in intergalactic space. Recent news relayed studies confirming cold gas ligaments between galaxy clusters, in which particles are very far apart, moving really slow and radiation is emitted which is interpreted as a gauge for the temperature (which is so weak, ie. cold, it wasn't detected in previous surveys).

edit: The point is, the far distance leaves the particles "alone" pretty much. If it doesn't make sense for a single paryicle to be seen as a closed system, of course it can't be assigned a temperature. But with regards to cooling and a strict loss of energy, the system is semipermeable, isolated in one direction. Then from our point of view, there's heat.

So for example, couldn't the temperature "be in" the electrons movement around the core (for lack of a better expression)? I mean, for a true resting state I would expect the electrons to fall into the core, not just ground state (annihilating the charges, so an atom really couldn't exist at absolut zero.

But if you are looking at a point source as the center of your frame of reference, it couldn't exhibit brownian motion. Then maybe curves described by spacetime has potential wells and exited states above that and maybe particles jitter between closely parallel worldlines, if they are entangled with anything besides.

[evanb]

It seems like you're confused about zero-temperature quantum mechanics. Even at absolute zero, the electrons do not 'fall into the core'---they sit in the orbitals---except at extraordinarily high pressures/densities where (as in neutron stars) they merge with the protons, turn into neutrons, emitting a neutrino.

Note that neutrons are heavier than protons, and when alone undergo beta decay with a lifetime of about 15 minutes, turning into a proton, emitting an electron and an antineutrino. So a hydrogen atom, left alone, cannot undergo the process you suggest---there simply isn't enough energy for it. It's only at very high density that the coulomb interaction provides enough energy to perform this conversion. There may be new physics, such as proton decay, that changes this story, but as far as anybody knows, there is not.

Perhaps more precisely, at absolute zero temperature atoms with orbitals are mathematically what quantum mechanics predicts. I of course don't know how to produce an atom that cold in the lab :) Except... of course... I do know how to produce an atom where the electron is in its ground state. If I've just got one atom, that's easy-peasy. Then, what is its temperature? Because it's just one atom, it's not well defined.

There's a more interesting question you ask, though: can the electron have a tempature? The different energy eigenstates are populated with different amplitudes. The way those states are populated can have a well-defined distribution in terms of a superposition. But if we restrict ourselves to a classical case, as OP was thinking about, superpositions are not allowed. Then, again, you at least have a definite energy.

Moreover, just a single electron in a superposition of energy eigenstates... it's not clear to me that that thing has a well-defined temperature either, because 'average' energy means something other than an ensemble average.

[posterboy]

> at absolute zero, the electrons do not 'fall into the core'--

well, actually I think nobody has shown this empirically. To add to my point, I don't think there can't be any closed system but one and it seems reasonable to assume that any substem could be at zero only if all are. Maybe energy just cant retrivably convert to other forms and loose all heat in the process when it's environment is already at rest. At best you could maybe get singularities before the environment exited by the process feeds heat back into the subsystem. Anything else would be as good as a papetuum mobile which you'd need to cool down this far.